A hyperbola has its centre at the origin, passes through the point (4, 2) and has transverse axis of length 4 along the x-axis. Then the eccentricity of the hyperbola is

Options

A. `(2)/(sqrt(3))`
B. `(3)/(2)`
C. `sqrt(3)`
D. 2

Answer

To solve the problem step by step, we will follow the information given about the hyperbola and use the standard form of the hyperbola equation.

### Step-by-Step Solution

1. **Identify the Standard Form of the Hyperbola**:
Since the hyperbola has its center at the origin and the transverse axis along the x-axis, the standard form of the hyperbola is given by:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

2. **Determine the Value of \( a \)**:
The problem states that the length of the transverse axis is 4. Therefore, we have:
\[ 2a = 4 \implies a = 2 \]
Thus, \( a^2 = 2^2 = 4 \).

3. **Substitute the Point (4, 2)**:
The hyperbola passes through the point (4, 2). We will substitute \( x = 4 \) and \( y = 2 \) into the hyperbola equation:
\[ \frac{4^2}{4} - \frac{2^2}{b^2} = 1 \]

4. **Calculate the Left Side**:
Calculate \( \frac{4^2}{4} \):
\[ \frac{16}{4} = 4 \]
Now substitute this into the equation:
\[ 4 - \frac{4}{b^2} = 1 \]

5. **Solve for \( b^2 \)**:
Rearranging the equation gives:
\[ 4 - 1 = \frac{4}{b^2} \implies 3 = \frac{4}{b^2} \]
Cross-multiplying results in:
\[ 3b^2 = 4 \implies b^2 = \frac{4}{3} \]

6. **Calculate the Eccentricity \( e \)**:
The eccentricity \( e \) of a hyperbola is given by the formula:
\[ e = \sqrt{1 + \frac{b^2}{a^2}} \]
Substituting \( b^2 = \frac{4}{3} \) and \( a^2 = 4 \):
\[ e = \sqrt{1 + \frac{\frac{4}{3}}{4}} = \sqrt{1 + \frac{4}{12}} = \sqrt{1 + \frac{1}{3}} = \sqrt{\frac{4}{3}} \]

7. **Final Calculation**:
Simplifying gives:
\[ e = \frac{2}{\sqrt{3}} \]

### Conclusion:
The eccentricity of the hyperbola is:
\[ \boxed{\frac{2}{\sqrt{3}}} \]